MATH 304
Linear Algebra
Lecture 13:
Span. Spanning set.
Subspaces of vector spaces
Definition. A vector space V0 is a subspace of a
vector space V if V0 ⊂ V and the linear operations
on V0 agree with the linear operations on V .
Proposition A subset S of a vector space V is a
subspace of V if and only if S is nonempty and
closed under linear operations, i.e.,
x, y ∈ S =⇒ x + y ∈ S,
x ∈ S =⇒ r x ∈ S for all r ∈ R.
Remarks. The zero vector in a subspace is the
same as the zero vector in V . Also, the subtraction
in a subspace agrees with that in V .
Examples of subspaces
• F (R): all functions f : R → R
• C (R): all continuous functions f : R → R
C (R) is a subspace of F (R).
• P: polynomials p(x) = a0 + a1 x + · · · + an−1 x n−1
• Pn : polynomials of degree less than n
Pn is a subspace of P.
• Any vector space V
• {0}, where 0 is the zero vector in V
The trivial space {0} is a subspace of V .
System of linear equations:

a11x1 + a12x2 + · · · + a1n xn = b1



a21x1 + a22x2 + · · · + a2n xn = b2
·········



am1 x1 + am2 x2 + · · · + amn xn = bm
Any solution (x1, x2, . . . , xn ) is an element of Rn .
Theorem The solution set of the system is a
subspace of Rn if and only if all bi = 0.
Examples of subspaces of M2,2(R): A =
• diagonal matrices: b = c = 0
• upper triangular matrices: c = 0
• lower triangular matrices: b = 0
• symmetric matrices (AT = A): b = c
• anti-symmetric matrices (AT = −A):
a = d = 0 and c = −b
• matrices with zero trace: a + d = 0
(trace = the sum of diagonal entries)
a b
c d
Let V be a vector space and v1 , v2, . . . , vn ∈ V .
Consider the set L of all linear combinations
r1 v1 + r2 v2 + · · · + rn vn , where r1, r2, . . . , rn ∈ R.
Theorem L is a subspace of V .
Proof: First of all, L is not empty. For example,
0 = 0v1 + 0v2 + · · · + 0vn belongs to L.
The set L is closed under addition since
(r1 v1 +r2 v2 + · · · +rn vn ) + (s1 v1 +s2 v2 + · · · +sn vn ) =
= (r1 +s1 )v1 + (r2 +s2 )v2 + · · · + (rn +sn )vn .
The set L is closed under scalar multiplication since
t(r1 v1 +r2 v2 + · · · +rn vn ) = (tr1 )v1 +(tr2 )v2 + · · · +(trn )vn .
Span: implicit definition
Let S be a subset of a vector space V .
Definition. The span of the set S, denoted
Span(S), is the smallest subspace of V that
contains S. That is,
• Span(S) is a subspace of V ;
• for any subspace W ⊂ V one has
S ⊂ W =⇒ Span(S) ⊂ W .
Remark. The span of any set S ⊂ V is well
defined (it is the intersection of all subspaces of V
that contain S).
Span: effective description
Let S be a subset of a vector space V .
• If S = {v1, v2, . . . , vn } then Span(S) is the set
of all linear combinations r1v1 + r2v2 + · · · + rn vn ,
where r1, r2, . . . , rn ∈ R.
• If S is an infinite set then Span(S) is the set of
all linear combinations r1 u1 + r2u2 + · · · + rk uk ,
where u1, u2 , . . . , uk ∈ S and r1 , r2, . . . , rk ∈ R
(k ≥ 1).
• If S is the empty set then Span(S) = {0}.
Examples of subspaces of M2,2(R):
0 0
1 0
and
consists of all
• The span of
0 0
0 1
matrices of the
form
a
1 0
0 0
+b
0 0
0 1
=
a 0
.
0 b
This is the subspace of diagonal matrices.
1 0
0 0
0 1
• The span of
,
, and
0 0
0 1
1 0
consistsof all matrices
of
theform a
1 0
0 0
+b
0 0
0 1
+c
0 1
1 0
=
a c
.
c b
This is the subspace of symmetric matrices.
Examples of subspaces of M2,2(R):
0 −1
• The span of
is the subspace of
1 0
anti-symmetric matrices.
1 0
0 0
0 1
• The span of
,
, and
0 0
0 1
0 0
is the subspace of upper triangular matrices.
1 0
0 0
0 1
0 0
• The span of
,
,
,
0 0
0 1
0 0
1 0
is the entire space M2,2(R).
Spanning set
Definition. A subset S of a vector space V is
called a spanning set for V if Span(S) = V .
Examples.
• Vectors e1 = (1, 0, 0), e2 = (0, 1, 0), and
e3 = (0, 0, 1) form a spanning set for R3 as
(x, y , z) = xe1 + y e2 + ze3 .
1 0
0 1
0 0
0 0
• Matrices
,
,
,
0 0
0 0
1 0
0 1
form a spanning set for M2,2(R) as
a b
c d
1 0
0 1
0 0
0 0
=a
+b
+c
+d
.
0 0
0 0
1 0
0 1
Problem Let v1 = (1, 2, 0), v2 = (3, 1, 1), and
w = (4, −7, 3). Determine whether w belongs to
Span(v1, v2).
We have to check if there exist r1, r2 ∈ R such that
w = r1v1 + r2v2. This vector equation is equivalent
to a system of linear equations:

 4 = r1 + 3r2
r1 = −5
−7 = 2r1 + r2 ⇐⇒
r2 = 3

3 = 0r1 + r2
Thus w = −5v1 + 3v2 is in Span(v1, v2).
Problem Let v1 = (2, 5) and v2 = (1, 3). Show
that {v1 , v2} is a spanning set for R2 .
Take any vector w = (a, b) ∈ R2 . We have to
check that there exist r1, r2 ∈ R such that
2r1 + r2 = a
w = r1v1 +r2v2 ⇐⇒
5r1 + 3r2 = b
2 1
Coefficient matrix: C =
. det C = 1 6= 0.
5 3
Since the matrix C is invertible, the system has a
unique solution for any a and b.
Thus Span(v1, v2) = R2 .
Problem Let v1 = (2, 5) and v2 = (1, 3). Show
that {v1 , v2} is a spanning set for R2 .
Alternative solution: First let us show that vectors
e1 = (1, 0) and e2 = (0, 1) belong to Span(v1, v2).
2r1 + r2 = 1
r1 = 3
e1 = r1 v1+r2 v2 ⇐⇒
⇐⇒
5r1 + 3r2 = 0
r2 = −5
2r1 + r2 = 0
r1 = −1
e2 = r1 v1+r2 v2 ⇐⇒
⇐⇒
5r1 + 3r2 = 1
r2 = 2
Thus e1 = 3v1 − 5v2 and e2 = −v1 + 2v2.
Then for any vector w = (a, b) ∈ R2 we have
w = ae1 + be2 = a(3v1 − 5v2) + b(−v1 + 2v2)
= (3a − b)v1 + (−5a + 2b)v2.
Problem Let v1 = (2, 5) and v2 = (1, 3). Show
that {v1 , v2} is a spanning set for R2 .
Remarks on the alternative solution:
Notice that R2 is spanned by vectors e1 = (1, 0)
and e2 = (0, 1) since (a, b) = ae1 + be2 .
This is why we have checked that vectors e1 and e2
belong to Span(v1, v2). Then
e1 , e2 ∈ Span(v1, v2) =⇒ Span(e1 , e2) ⊂ Span(v1, v2)
=⇒ R2 ⊂ Span(v1, v2) =⇒ Span(v1, v2) = R2 .
In general, to show that Span(S1) = Span(S2),
it is enough to check that S1 ⊂ Span(S2 ) and
S2 ⊂ Span(S1).
More properties of span
Let S0 and S be subsets of a vector space V .
• S0 ⊂ S =⇒ Span(S0) ⊂ Span(S).
• Span(S0 ) = V and S0 ⊂ S =⇒ Span(S) = V .
• If v0 , v1, . . . , vk is a spanning set for V and v0
is a linear combination of vectors v1 , . . . , vk then
v1 , . . . , vk is also a spanning set for V .
Indeed, if v0 = r1 v1 + · · · + rk vk , then
t0 v0 + t1 v1 + · · · + tk vk = (t0 r1 + t1 )v1 + · · · + (t0 rk + tk )vk .
• Span(S0 ∪ {v0}) = Span(S0 ) if and only if
v0 ∈ Span(S0 ).
If v0 ∈ Span(S0 ), then S0 ∪ v0 ⊂ Span(S0 ), which implies
Span(S0 ∪ {v0 }) ⊂ Span(S0 ). On the other hand,
Span(S0 ) ⊂ Span(S0 ∪ {v0 }).